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Standard X

Chemistry

Effect on Indicators by Acids and Bases

Phenolphthale...

Question

Phenolphthalein indicator is colourless in an acid and turns pink in an alkaline solution. When an aqueous solution of potassium iodide is electrolysed with phenolphthale in indicator, what colours appear at the electrodes?
Cathode: $2 H_{2} O_{(l)} + 2 e^{-} \to H_{(g)} + 2 O H^{-} (a q)$
Anode: $2 I_{(a q)}^{-} \to I_{2 (s)} + 2 e^{-}$

Brown at the cathode and pink at the anode

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Red at the cathode and pink at the anode

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Brown at the anode and colourless at the cathode

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Pink at the cathode and brown at the anode

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Solution

The correct option is C Pink at the cathode and brown at the anode
Cathode: $2 H_{2} O + 2 e^{-} ⟶ H (g) + 2 O H^{-}$
Here $O H^{-}$ is alkaline so phenolphtale is turned pink.
Anode: $2 I^{'} ⟶ I_{2} (s) + 2 e^{-}$
$I^{-} + I_{2} ⟶ I_{3}$
This is the reason for brown colour in anode.

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Phenolphthalein indicator is colourless in an acid and turns pink in an alkaline solution. When an aqueous solution of potassium iodide is electrolysed with phenolphthale in indicator, what colours appear at the electrodes?Cathode: 2H2O(l)+2e−→H(g)+2OH−(aq)Anode: 2I−(aq)→I2(s)+2e−

The correct option is C Pink at the cathode and brown at the anodeCathode: 2H2O+2e−⟶H(g)+2OH−Here OH− is alkaline so phenolphtale is turned pink.Anode: 2I′⟶I2(s)+2e−I−+I2⟶I3This is the reason for brown colour in anode.

The correct option is C Pink at the cathode and brown at the anode
Cathode: $2 H_{2} O + 2 e^{-} ⟶ H (g) + 2 O H^{-}$
Here $O H^{-}$ is alkaline so phenolphtale is turned pink.
Anode: $2 I^{'} ⟶ I_{2} (s) + 2 e^{-}$
$I^{-} + I_{2} ⟶ I_{3}$
This is the reason for brown colour in anode.